I am trying to understand Exercise 1.10 in Rudin, reproduced below. The part that I am having difficult with is bolded.
Suppose $z = a + bi$, $w = u + iv$ and $a = \left(\frac{|w| + u}{2}\right)^{1/2}$, $b = \left(\frac{|w| - 2}{2}\right)^{1/2}$. Prove that $z^2 = w$ if $v \geq 0$ and that $\left(\overline{z}\right)^2 = w$ if $v \leq 0$. Conclude that every complex number (with one exception) has two complex square roots.
With some algebra, I have successfully proved that $z^2 = w$ if $v \geq 0$ and $\left(\overline{z}\right)^2 = w$ if $v \leq 0$. The next statement should be rather obvious, but I cannot see it. The exception seems to be the case $z = 0$, I assume because $0 = -0$. However, I cannot see directly how that follows from this above chain of logic. Rudin's solution seem to deduce this final line upon proving the above without explaining why it follows.
Any help on this would be greatly appreciated.